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How would I integrate different values into 4 tables?

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I have this code
clc
clear
close
a = 40*(pi/180);
v = 1200;
g = 32.2;
k=[0,2e-6,10e-6,20e-6];
k=0;
dt=.5;
t = 0:dt:55;
v = zeros(length(t),4);
vx = zeros(length(t),4);
vy= zeros(length(t),4);
x = zeros(length(t),4);
y = zeros(length(t),4);
v(1,1) = 1200;
vx(1,1) = v(1,1)*cos(a);
vy(1,1) = v(1,1)*sin(a);
x(1,1)=0;
y(1,1)=0;
for i=2:length (t)
x(i,1) = x(i-1,1)+vx(i-1,1)*dt-.5*k*v(i-1,1).^2*cos(a)*dt.^2;
y(i,1) = y(i-1,1)+vy(i-1,1)*dt-.5*k*v(i-1,1).^2*sin(a)*dt^2-0.5*g*dt.^2;
vx(i,1) = vx(i-1,1)-k*v(i-1,1).^2*cos(a)*dt;
vy(i,1) = vy(i-1,1)-k*v(i-1,1).^2*sin(a)*dt-g*dt;
a=atan(vy(i,1)./vx(i,1));
v(i,1)=sqrt(vx(i,1).^2+vy(i,1).^2);
end
vx(1:round(5/dt):end,:)
vy(1:round(5/dt):end,:)
x(1:round(5/dt):end,:)
y(1:round(5/dt):end,:)
That results in
ans =
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
919.25 0 0 0
ans =
771.35 0 0 0
610.35 0 0 0
449.35 0 0 0
288.35 0 0 0
127.35 0 0 0
-33.65 0 0 0
-194.65 0 0 0
-355.65 0 0 0
-516.65 0 0 0
-677.65 0 0 0
-838.65 0 0 0
-999.65 0 0 0
ans =
0 0 0 0
4596.27 0 0 0
9192.53 0 0 0
13788.80 0 0 0
18385.07 0 0 0
22981.33 0 0 0
27577.60 0 0 0
32173.87 0 0 0
36770.13 0 0 0
41366.40 0 0 0
45962.67 0 0 0
50558.93 0 0 0
ans =
0 0 0 0
3454.23 0 0 0
6103.45 0 0 0
7947.68 0 0 0
8986.90 0 0 0
9221.13 0 0 0
8650.35 0 0 0
7274.58 0 0 0
5093.81 0 0 0
2108.03 0 0 0
-1682.74 0 0 0
-6278.52 0 0 0
My issues is whenever I change the K value to [2e-6,10e-6, or 20e-6] it results in the first column changing to the respective k values. However, if I change the x(i,1) positions to say x(i,2) for [2e-6] (and all subsequent notations) then I get every k = 2e-6 value in the 2nd column and the first column becomes zero. Question is how do I get the answers to add on to the table made from previous calculations?

  2 Comments

BALAJI KARTHEEK
BALAJI KARTHEEK on 20 Apr 2020
Just give sample image of the table (what ur expecting in the result), so it will easy to understand and modify the code written by u..
Michael Sabol
Michael Sabol on 20 Apr 2020
The answer to the above code is listed. I would like to have the three other columns filled with the coressponding k values (2e-6, 10e-6, and 20e-6)

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Accepted Answer

BALAJI KARTHEEK
BALAJI KARTHEEK on 21 Apr 2020
I modified the code for you and i posted the ans also, i hope this is what u wanted.....
clc
clear
close
format long
k1=[0,2e-6,10e-6,20e-6];
dt=.5;
t = 0:dt:55;
g = 32.2;
v = zeros(length(t),4);
vx = zeros(length(t),4);
vy= zeros(length(t),4);
x = zeros(length(t),4);
y = zeros(length(t),4);
for j=1:length(k1)
a = 40*(pi/180);
k=k1(j);
v(1,j) = 1200;
vx(1,j) = v(1,j)*cos(a);
vy(1,j) = v(1,j)*sin(a);
x(1,j)=0;
y(1,j)=0;
for i=2:length (t)
x(i,j) = x(i-1,j)+vx(i-1,j)*dt-.5*k*v(i-1,j).^2*cos(a)*dt.^2;
y(i,j) = y(i-1,j)+vy(i-1,j)*dt-.5*k*v(i-1,j).^2*sin(a)*dt^2-0.5*g*dt.^2;
vx(i,j) = vx(i-1,j)-k*v(i-1,j).^2*cos(a)*dt;
vy(i,j) = vy(i-1,j)-k*v(i-1,j).^2*sin(a)*dt-g*dt;
a=atan(vy(i,j)./vx(i,j));
v(i,j)=sqrt(vx(i,j).^2+vy(i,j).^2);
end
end
vx(1:round(5/dt):end,:)
vy(1:round(5/dt):end,:)
x(1:round(5/dt):end,:)
y(1:round(5/dt):end,:)
ans =
1.0e+02 *
9.192533317427737 9.192533317427737 9.192533317427737 9.192533317427737
9.192533317427737 9.087437198565105 8.688194539448844 8.232369634405721
9.192533317427737 8.992683198820997 8.271606474129895 7.515600132134569
9.192533317427737 8.906142118918728 7.919402869533692 6.956903130394356
9.192533317427737 8.825438093191325 7.611669801834986 6.497663684523137
9.192533317427737 8.747994917296394 7.330900766229695 6.094617348287314
9.192533317427737 8.671191392627970 7.062225373628980 5.716607364427984
9.192533317427737 8.592588049635696 6.794295817589458 5.344401073260919
9.192533317427737 8.510130827164266 6.519858929429604 4.969536950219919
9.192533317427737 8.422255391440707 6.235515863016696 4.591412572439150
9.192533317427737 8.327886941553478 5.940899030969548 4.213955536697458
9.192533317427737 8.226377828143628 5.637702715980655 3.842942121374995
ans =
1.0e+02 *
7.713451316238471 7.713451316238471 7.713451316238471 7.713451316238471
6.103451316238469 6.023419388685378 5.719372277363858 5.372185731794696
4.493451316238467 4.358060751416640 3.869349418126093 3.356497018654969
2.883451316238464 2.713013480784294 2.125068241342515 1.550149352309664
1.273451316238464 1.084948020280437 0.460462096125176 -0.114597066820229
-0.336548683761536 -0.528223533354256 -1.139760914385037 -1.672277710014929
-1.946548683761535 -2.127194956892446 -2.681175666264873 -3.132854106675353
-3.556548683761536 -3.711278137353608 -4.161511360325129 -4.490534191556887
-5.166548683761538 -5.278613302973931 -5.573538912210908 -5.733287819377662
-6.776548683761540 -6.826502519553036 -6.908138272916148 -6.850276415575496
-8.386548683761543 -8.351753997529265 -8.156695586961243 -7.835785114702563
-9.996548683761544 -9.850966102443524 -9.312567227724861 -8.690247541356209
ans =
1.0e+04 *
0 0 0 0
0.459626665871387 0.456952509660115 0.446601570599481 0.434388168668790
0.919253331742773 0.908917067719421 0.870286135308686 0.827286469285742
1.378879997614160 1.356358536870645 1.274838854896066 1.188582146706933
1.838506663485547 1.799629168051811 1.662970072505337 1.524634799788792
2.298133329356934 2.238956995531161 2.036456085149122 1.839283305588884
2.757759995228320 2.674439272489372 2.396260757987131 2.134509243829672
3.217386661099707 3.106045768985768 2.742688806528300 2.411035206651566
3.677013326971094 3.533633248951142 3.075579187877789 2.668900501599922
4.136639992842481 3.956967824055119 3.394507100925129 2.907931858633309
4.596266658713867 4.375749794739506 3.698957566875462 3.128051985769669
5.055893324585254 4.789636722968088 3.988452648518830 3.329435350798285
ans =
1.0e+04 *
0 0 0 0
0.345422565811923 0.343309821893089 0.335128870726984 0.325469355578336
0.610345131623847 0.602754750055932 0.574341837338965 0.542628698537214
0.794767697435771 0.779455549117251 0.723828388470866 0.664600119087606
0.898690263247694 0.874339858815437 0.788177334077686 0.699997622986476
0.922112829059617 0.888199047586014 0.770947619004973 0.654920194605361
0.865035394871541 0.821754585219541 0.675181896918930 0.534387622251712
0.727457960683465 0.675728278727461 0.503850396003435 0.343355037091190
0.509380526495388 0.450906840766734 0.260173936315252 0.087260126553497
0.210803092307311 0.148192401693298 -0.052206230259897 -0.227864175692256
-0.168274341880766 -0.231364338893490 -0.429197553309034 -0.595561742932445
-0.627851776068843 -0.686546835784460 -0.866321819027821 -1.009244632798958

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More Answers (1)

darova
darova on 20 Apr 2020
Try this solution

  2 Comments

Michael Sabol
Michael Sabol on 20 Apr 2020
Hello thanks for the answer, one question. is the k=K1(j); inside the for loop you gave or my original one?
like so?
for j=1:length (k)
for i=2:length (t)
k=K1(j);
x(i,j) = x(i-1,j)+vx(i-1,j)*dt-.5*k*v(i-1,j).^2*cos(a)*dt.^2;
y(i,j) = y(i-1,j)+vy(i-1,j)*dt-.5*k*v(i-1,j).^2*sin(a)*dt^2-0.5*g*dt.^2;
vx(i,j) = vx(i-1,j)-k*v(i-1,j).^2*cos(a)*dt;
vy(i,j) = vy(i-1,j)-k*v(i-1,j).^2*sin(a)*dt-g*dt;
end
end
darova
darova on 20 Apr 2020
It doesn't matter. More reasonable be inside first
for j=1:length (k)
k=K1(j);
for i=2:length (t)
% code

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